Find the Inverse fifth root of 4x+2

$\sqrt[5]{4 x + 2}$

Step 1

Interchange the variables.

$x = \sqrt[5]{4 y + 2}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[5]{4 y + 2} = x$ .

$\sqrt[5]{4 y + 2} = x$

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $5$ .

${\sqrt[5]{4 y + 2}}^{5} = x^{5}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{4 y + 2}$ as ${(4 y + 2)}^{\frac{1}{5}}$ .

${({(4 y + 2)}^{\frac{1}{5}})}^{5} = x^{5}$

Simplify the left side.

Simplify ${({(4 y + 2)}^{\frac{1}{5}})}^{5}$ .

Multiply the exponents in ${({(4 y + 2)}^{\frac{1}{5}})}^{5}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(4 y + 2)}^{\frac{1}{5} \cdot 5} = x^{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

${(4 y + 2)}^{\frac{1}{5} \cdot 5} = x^{5}$

Rewrite the expression.

${(4 y + 2)}^{1} = x^{5}$

Simplify.

$4 y + 2 = x^{5}$

Solve for $y$ .

Subtract $2$ from both sides of the equation.

$4 y = x^{5} - 2$

Divide each term in $4 y = x^{5} - 2$ by $4$ and simplify.

Divide each term in $4 y = x^{5} - 2$ by $4$ .

$\frac{4 y}{4} = \frac{x^{5}}{4} + \frac{- 2}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{x^{5}}{4} + \frac{- 2}{4}$

Divide $y$ by $1$ .

$y = \frac{x^{5}}{4} + \frac{- 2}{4}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 2$ and $4$ .

Factor $2$ out of $- 2$ .

$y = \frac{x^{5}}{4} + \frac{2 (- 1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$y = \frac{x^{5}}{4} + \frac{2 \cdot - 1}{2 \cdot 2}$

Cancel the common factor.

$y = \frac{x^{5}}{4} + \frac{2 \cdot - 1}{2 \cdot 2}$

Rewrite the expression.

$y = \frac{x^{5}}{4} + \frac{- 1}{2}$

Move the negative in front of the fraction.

$y = \frac{x^{5}}{4} - \frac{1}{2}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{x^{5}}{4} - \frac{1}{2}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{5}}{4} - \frac{1}{2}$ is the inverse of $f (x) = \sqrt[5]{4 x + 2}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\sqrt[5]{4 x + 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{(\sqrt[5]{4 x + 2})}^{5}}{4} - \frac{1}{2}$

Simplify each term.

Simplify the numerator.

Factor $2$ out of $4 x + 2$ .

Factor $2$ out of $4 x$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{\sqrt[5]{2 (2 x) + 2}}^{5}}{4} - \frac{1}{2}$

Factor $2$ out of $2$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{\sqrt[5]{2 (2 x) + 2 (1)}}^{5}}{4} - \frac{1}{2}$

Factor $2$ out of $2 (2 x) + 2 (1)$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{\sqrt[5]{2 (2 x + 1)}}^{5}}{4} - \frac{1}{2}$

Rewrite ${\sqrt[5]{2 (2 x + 1)}}^{5}$ as $2 (2 x + 1)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{2 (2 x + 1)}$ as ${(2 (2 x + 1))}^{\frac{1}{5}}$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{({(2 (2 x + 1))}^{\frac{1}{5}})}^{5}}{4} - \frac{1}{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{(2 (2 x + 1))}^{\frac{1}{5} \cdot 5}}{4} - \frac{1}{2}$

Combine $\frac{1}{5}$ and $5$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{(2 (2 x + 1))}^{\frac{5}{5}}}{4} - \frac{1}{2}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{{(2 (2 x + 1))}^{\frac{5}{5}}}{4} - \frac{1}{2}$

Rewrite the expression.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x + 1)}{4} - \frac{1}{2}$

Simplify.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x + 1)}{4} - \frac{1}{2}$

Apply the distributive property.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x) + 2 \cdot 1}{4} - \frac{1}{2}$

Multiply $2$ by $2$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{4 x + 2 \cdot 1}{4} - \frac{1}{2}$

Multiply $2$ by $1$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{4 x + 2}{4} - \frac{1}{2}$

Factor $2$ out of $4 x + 2$ .

Factor $2$ out of $4 x$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x) + 2}{4} - \frac{1}{2}$

Factor $2$ out of $2$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x) + 2 (1)}{4} - \frac{1}{2}$

Factor $2$ out of $2 (2 x) + 2 (1)$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x + 1)}{4} - \frac{1}{2}$

Cancel the common factors.

Factor $2$ out of $4$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x + 1)}{2 \cdot 2} - \frac{1}{2}$

Cancel the common factor.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 (2 x + 1)}{2 \cdot 2} - \frac{1}{2}$

Rewrite the expression.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 x + 1}{2} - \frac{1}{2}$

Simplify terms.

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 x + 1 - 1}{2}$

Combine the opposite terms in $2 x + 1 - 1$ .

Subtract $1$ from $1$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt[5]{4 x + 2}) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt[5]{4 x + 2}) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{x^{5}}{4} - \frac{1}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{4 (\frac{x^{5}}{4} - \frac{1}{2}) + 2}$

Factor $2$ out of $4 (\frac{x^{5}}{4} - \frac{1}{2}) + 2$ .

Factor $2$ out of $4 (\frac{x^{5}}{4} - \frac{1}{2})$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (2 (\frac{x^{5}}{4} - \frac{1}{2})) + 2}$

Factor $2$ out of $2$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (2 (\frac{x^{5}}{4} - \frac{1}{2})) + 2 (1)}$

Factor $2$ out of $2 (2 (\frac{x^{5}}{4} - \frac{1}{2})) + 2 (1)$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (2 (\frac{x^{5}}{4} - \frac{1}{2}) + 1)}$

Apply the distributive property.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (2 (\frac{x^{5}}{4}) + 2 (- \frac{1}{2}) + 1)}$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (2 (\frac{x^{5}}{2 (2)}) + 2 (- \frac{1}{2}) + 1)}$

Cancel the common factor.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (2 (\frac{x^{5}}{2 \cdot 2}) + 2 (- \frac{1}{2}) + 1)}$

Rewrite the expression.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (\frac{x^{5}}{2} + 2 (- \frac{1}{2}) + 1)}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (\frac{x^{5}}{2} + 2 (\frac{- 1}{2}) + 1)}$

Cancel the common factor.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (\frac{x^{5}}{2} + 2 (\frac{- 1}{2}) + 1)}$

Rewrite the expression.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (\frac{x^{5}}{2} - 1 + 1)}$

Simplify by adding numbers.

Add $- 1$ and $1$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (\frac{x^{5}}{2} + 0)}$

Add $\frac{x^{5}}{2}$ and $0$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{2 (\frac{x^{5}}{2})}$

Combine $2$ and $\frac{x^{5}}{2}$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{\frac{2 x^{5}}{2}}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{2 x^{5}}{2}$ by cancelling the common factors.

Cancel the common factor.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{\frac{2 x^{5}}{2}}$

Rewrite the expression.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{\frac{x^{5}}{1}}$

Divide $x^{5}$ by $1$ .

$f (\frac{x^{5}}{4} - \frac{1}{2}) = \sqrt[5]{x^{5}}$

Pull terms out from under the radical, assuming real numbers.

$f (\frac{x^{5}}{4} - \frac{1}{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{5}}{4} - \frac{1}{2}$ is the inverse of $f (x) = \sqrt[5]{4 x + 2}$ .

$f^{- 1} (x) = \frac{x^{5}}{4} - \frac{1}{2}$

Find the Inverse fifth root of 4x+2

Related Questions

Support LCMGCF.com by disabling your adblocker.